Equilateral Triangle Base-Height Rationality

@Muhammad Rasel Parvej – Sure. $\mathbb Q(\sqrt 3)$ consists precisely of all numbers of the form $q_0 + q_1\sqrt 3$ , with rational $q_0,q_1$ .

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In general, if $R$ is a field and $x$ an element not in the field, $R(x)$ stands for the field of all fractions of polynomials generated by $x$ over $R$ , i.e. $R(x) = \left\{ \frac{p_0 + p_1x + p_2x^2 + \cdots + p_nx^n}{q_0 + q_1x + q_2x^2 + \cdots + q_mx^m}\ \text{with}\ p_i,q_i \in R\right\}.$ If $x$ is algebraic over $R$ of order $n$ , this field is finitely generated, since powers $x^n$ and greater may be reduced using the defining polynomial. Thus, if $x$ is quadratic over $R$ , we have $R(x) = \left\{ r_0 + r_1x,\ \ r_0,r_1 \in R\right\}.$

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One can add more than one element to a field in this fashion, e.g. $\mathbb Q(\sqrt 2)(\sqrt 3) = \mathbb Q(\sqrt 2, \sqrt 3)$ consists of all expressions of the form $q_0 + q_1\sqrt 2 + q_2\sqrt 3 + q_3\sqrt 6$ .

By continuing this process, one can define $\mathbb Q^{\sqrt{\ }}$ as the limit of including all quadratic irrationals, and eventually by adding the roots of all rational polynomials, one obtains $\mathbb Q^a$ , the algebraic closure of $\mathbb Q$ , which contains all algebraic numbers.

(Note: The set $\mathbb Q^{\sqrt{\ }} \cap \mathbb R^+$ is precisely the set of lengths that may be construed with straightedge and compass starting from a line segment of length 1.)

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In the same fashion, we may define for instance, $\mathbb Q^a(\pi)$ , which consists of all numbers that can be obtained by addition, subtraction, multiplication and division of rational or algebraic numbers and $\pi$ ; or $(\mathbb Q(\pi))^a$ , which also includes numbers such as $\sqrt \pi$ .